\documentclass[english]{article}
\usepackage{setspace}
\usepackage{babel}
\usepackage{amsmath}
\usepackage{amsthm}

\setlength{\oddsidemargin}{0in}
\setlength{\evensidemargin}{0in}
\setlength{\topmargin}{0in}
\setlength{\textheight}{8.5in}
\setlength{\textwidth}{6.5in}
\setlength{\parskip}{0.1in}
\setlength{\parindent}{0in}

\newtheorem*{thm}{Theorem}

\title{Notebook Problem 5.6.34}
\author{Jason Wojciechowski}

\begin{document}
\maketitle

\begin{thm} Let $f(x) \in F[x]$ where $F$ is a field, and let
  $\alpha \in F$. Then the remainder $r(x)$ when $f(x)$ is divided by
  $x-\alpha$ is $f(\alpha)$.
\end{thm}

\begin{proof} First, if $x - \alpha$ divides $f(x)$, then $\alpha$ is a
  zero of $f(x)$, so $f(\alpha)=0$, and the remainder $r(x)$ when
  $f(x)$ is divided by $x-\alpha$ is also $0$.
  
  Now, if $x-\alpha$ does not divide $f(x)$, the remainder $r(x)$ must
  have degree less than the degree of $x-\alpha$ (by the Division
  Algorithm), so $r(x)$ must be a constant term.  Call it $k$.  Thus,
  $f(x) = (x-\alpha)q(x) + k$ and $f(\alpha) =
  (\alpha-\alpha)q(\alpha) + k$, so that $f(\alpha) = k = r(x)$.
\end{proof}

\end{document}